Answer
We show that the series
$\pi - \dfrac{{{\pi ^3}}}{{3!}} + \dfrac{{{\pi ^5}}}{{5!}} - \dfrac{{{\pi ^7}}}{{7!}} + \cdot\cdot\cdot$
converges to zero.
Five terms must be computed so that the approximation is within $0.01$ of zero.
Work Step by Step
From Table 2 we get
$\sin x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}} = x - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - \dfrac{{{x^7}}}{{7!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$
Thus, our original series can be represented by
$\sin \pi = \pi - \dfrac{{{\pi ^3}}}{{3!}} + \dfrac{{{\pi ^5}}}{{5!}} - \dfrac{{{\pi ^7}}}{{7!}} + \cdot\cdot\cdot$
Since $\sin \pi = 0$, the series
$\pi - \dfrac{{{\pi ^3}}}{{3!}} + \dfrac{{{\pi ^5}}}{{5!}} - \dfrac{{{\pi ^7}}}{{7!}} + \cdot\cdot\cdot$,
converges to zero.
Notice that the positive terms of this series are ${b_n} = \dfrac{{{\pi ^{2n + 1}}}}{{\left( {2n + 1} \right)!}}$.
Let $S = \sin \pi = \pi - \dfrac{{{\pi ^3}}}{{3!}} + \dfrac{{{\pi ^5}}}{{5!}} - \dfrac{{{\pi ^7}}}{{7!}} + \cdot\cdot\cdot$.
By Eq. (2) in Section 11.4:
$\left| {S - {S_n}} \right| \lt {b_{n + 1}}$
We choose $n$ so that
$\left| {S - {S_n}} \right| \lt {b_{n + 1}} \lt {10^{ - 2}}$
$\dfrac{{{\pi ^{2n + 3}}}}{{\left( {2n + 3} \right)!}} \lt {10^{ - 2}}$
$\dfrac{{\left( {2n + 3} \right)!}}{{{\pi ^{2n + 3}}}} \gt {10^2}$
We evaluate and list some values for $n$ in the following table:
$\begin{array}{*{20}{c}}
n&{}&{\dfrac{{\left( {2n + 3} \right)!}}{{{\pi ^{2n + 3}}}}}\\
2&{}&{1.67}\\
3&{}&{12.17}\\
4&{}&{135.68}\\
5&{}&{2144.53}
\end{array}$
Based on the results in the table above, we choose $N=4$, or five terms required such that the approximation is within $0.01$ of zero.
Using a calculator, we compute: ${S_4} \approx 0.0069253$.
Hence, $\left| {0 - {S_4}} \right| \approx 0.0069253 \lt 0.01$.
